Download the app

Questions  

A particle moving in the positive X direction has de Broglie wavelength λ1. Another particle also moving in the same direction has de Broglie wavelength λ2. The two particles make a collision and after the collision they move together as a single particle. The de Broglie wavelength of this single particle is given by 

a
λ1+λ22
b
λ1λ2
c
2λ1λ2λ1 +λ2
d
λ1λ2λ1 +λ2

detailed solution

Correct option is D

λ1 =hp1     p1 =hλ1       similarly p2= hλ2 after collision p =p1  + p2 λ =hp =λ1λ2λ1+λ2

Talk to our academic expert!

+91

Are you a Sri Chaitanya student?


Similar Questions

An electron is orbiting in 4th energy state in a hydrogen atom. The de-Broglie wavelength associated with this electron is nearly equal to


phone icon
whats app icon